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Question Number 131202 by abdurehime last updated on 02/Feb/21

$$1{\lim }_{\mathrm{n}\to \mathrm{\infty }}\left(\frac{3^{\mathrm{n}}+2^{\mathrm{n}}}{6^{\mathrm{n}}}\right)\mathrm{and}$$$$\stackrel{\lim \underset{\mathrm{n}\to \mathrm{\infty }}{}\left(\frac{1+2^2+32+4^2+.......+{\mathrm{n}}^2}{{\mathrm{n}}^3}\right)}{}$$$$\mathrm{help}\mathrm{me}$$

Commented by liberty last updated on 02/Feb/21

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left(\frac{1^2+2^2+3^2+4^2+...+{\mathrm{n}}^2}{{\mathrm{n}}^3}\right)$$

Answered by liberty last updated on 02/Feb/21

$$\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\frac{6^{\mathrm{n}}({\left(\frac{3}{6}\right)}^{\mathrm{n}}+{\left(\frac{2}{6}\right)}^{\mathrm{n}})}{6^{\mathrm{n}}}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }[{\left(\frac{1}{2}\right)}^{\mathrm{n}}+{\left(\frac{1}{3}\right)}^{\mathrm{n}}]$$$$=0+0=0$$

Commented by EDWIN88 last updated on 03/Feb/21

$$itisrightanswer!$$

Answered by Chhing last updated on 02/Feb/21

$$$$$$=\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{\frac{1}{6}\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{{\mathrm{n}}^3}\right)$$$$=\frac{1}{6}\underset{x\to \mathrm{\infty }}{\lim }\left(\frac{{\mathrm{n}}^3(2+\frac{3}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^2})}{{\mathrm{n}}^3}\right)$$$$=\frac{1}{6}\times \underset{x\to \mathrm{\infty }}{\lim }(2+\frac{3}{\mathrm{n}}+\frac{1}{{\mathrm{n}}^2})$$$$=\frac{1}{6}\times 2$$$$=\frac{1}{3}$$

Answered by Ar Brandon last updated on 02/Feb/21

$$1.\mathcal{L}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left(\frac{3^{\mathrm{n}}+2^{\mathrm{n}}}{6^{\mathrm{n}}}\right)=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\{{\left(\frac{3}{6}\right)}^{\mathrm{n}}+{\left(\frac{2}{6}\right)}^{\mathrm{n}}\}=0$$$$2.\mathcal{T}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left\{\frac{1+2^2+3^2+4^2+\cdot \cdot \cdot +{\mathrm{n}}^2}{{\mathrm{n}}^3}\right\}=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{{\mathrm{k}}^2}{{\mathrm{n}}^3}$$$$=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }\left\{\frac{1}{\mathrm{n}}\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{{\mathrm{k}}^2}{{\mathrm{n}}^2}\right\}={\int }_0^1{\mathrm{x}}^2\mathrm{dx}=\frac{1}{3}$$$$$$

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